No More Lower Arcs

 In the previous post "Looking For P4?" dated 02 Dec 2023, the arc \(P1+P2+2\) is not excluded from providing a \(q\).  This \(q=1\), and \(P3=P1+P2+2-2=P1+P2\).

And there is no need for full lower arcs, just long enough to provide the first integer marking.

This is how arc \(P1+P2-2\) provides a possible solution always,

as \(P1+P2+2\) is always even with \(P1\) and \(P2\) being prime and \(P1+P2-2\) is short by four integer lengths.  Therefore, two marking (one on each portion) on arc \(P1+P2-2\), will accommodate a line through it that intersect another integer marking on arc \(P1+P2+2\).

Possibly,

\(P4=P1-2\),     \(P3=P2+2\)

or

\(P4=P1+2\),     \(P3=P2-2\)

when the radial intersects arc \(P1+P2+2\) and recovers 4 integer units in total.